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As shown in the figure, it is known that in the parallelogram ABCD, e and F are respectively the midpoint of AD, G and H are two points on the diagonal BD, and BG = DH, what conclusion can be obtained?
Solution: △ GFB ≌ △ hed proves that: ∵ ABCD is a parallelogram ∵ 1 / 2BC = 1 / 2ad BF = ed ∵ BC / / ad ≌ CBD = ∠ BDA in △ GFB and △ hed, BG = DH ∠ CBD = ∠ BDA
As shown in the figure, in the parallelogram ABCD, be and DF are perpendicular to AC, e and f respectively. Suppose de equals BF? Try to explain the reason
The reasons are as follows: ∵ quadrilateral ABCD is parallelogram, ∵ ad = BC, ad ∥ BC, ∵ DAC = ∥ BCA, ∵ DF ⊥ AC, be ⊥ AC, ∵ AFD = ∥ CEB = ∥ DFE = ∥ AEB = 90 °, ∥ ADF ≌ BEC, ∥ DF = be, ∥ EF = EF, ≌ DFE ≌ bef, ∥ de = BF
It is known that, as shown in the figure, ▱ ABCD, points E and F are on CD and ab respectively, DF ‖ be and EF intersect BD at point o
The ∵ quadrilateral ABCD is a parallelogram, ∵ ab ∥ CD, ∵ DF ∥ be, ∵ quadrilateral BEDF is a parallelogram, ∥ OE = of
RELATED INFORMATIONS
- 1. As shown in the figure, in the parallelogram ABCD, points E and F are on CD and ab respectively, and DF / / BF is on the point
- 2. As shown in the figure, fold a square piece of paper ABCD along EF, AD / / BC. If the angle EFG = 52 °, calculate the degree of angle deg
- 3. As shown in the figure, after a rectangular piece of paper ABCD is folded along EF, points D and C respectively fall at d 'C', the intersection of ED 'and BC is g, if ∠ EFG = 65 degrees Find the degree of ∠ 1 and ∠ 2
- 4. When a rectangular piece of paper ABCD is folded along EF, the intersection points of ED and BC are g, D and C at M and N, respectively. If ∠ EFG = 55 °, then ∠ 1=______ °，∠2=______ °．
- 5. (Mianyang, 2011) as shown in the figure, fold a rectangular piece of paper ABCD with a length of 8cm and a width of 4cm so that point a and point C coincide, and the length of crease EF is equal to 2525cm
- 6. The edge length of rectangular paper ABCD is ab = 4, ad = 2. Fold the rectangular paper along EF so that point a and point C coincide. After folding, color it on one side (as shown in the figure), the area of the colored part is () A. 8B. 112C. 4D. 52
- 7. In the parallelogram ABCD, ab = KBC, point P is on the diagonal BD, and angle ∠ ape = ∠ bad, PE and BC are compared with point E, and the quantitative relationship between PA and PE is explored
- 8. As shown in the figure, in the parallelogram ABCD, e is the midpoint of AD, the intersection of the extension line of CE and Ba and the point F, if BD = 2CD, the verification: ∠ f = ∠ BCF
- 9. In the parallelogram ABCD, is the midpoint of e-type CD, a point on the edge of F-type BC, and the angle FAE = angle ead, EF and AE perpendicular? Give reasons
- 10. In the parallelogram ABCD, what is the length of CE if the extension line of AE bisector angle bad and intersecting DC is at point E, AB equals 4 and ad equals 9?
- 11. Parallelogram ABCD, AC = 2gc, AE = EF = FB, triangle GEF area is 6 square centimeter, find the area of parallelogram
- 12. The parallelogram ABCD, e, f are the trisection points of BC, connecting BD, AE, AF to intersect BD with m, N, finding: Mn: Nd It is known that: the parallelogram ABCD, f is the trisection point of BC, connecting BD, AE, AF with BD in M, and finding: Mn: Nd
- 13. It is known that the perimeter of the parallelogram ABCD is 32, ab = 4, then BC is equal to
- 14. As shown in the figure, the diagonals AC and BD of rectangular ABCD are called o, and E F are the midpoint of OA and ob respectively If ad = 4cm, ab = 8cm, find the length of DF of EF
- 15. As shown in the figure, the diagonals AC and BD of rectangle ABCD intersect at O, e and F are the midpoint of OA and ob respectively. (1) prove: △ ade ≌ △ BCF; (2) if ad = 4cm, ab = 8cm, find the length of CF
- 16. If vector AB = vector DC, then ABCD is a parallelogram? Is that right? Maybe it's not collinear
- 17. It is known that in the quadrilateral ABCD, ab = ad = 4, BC = 6, CD = 2,3 vector AB * vector AD + 4 vector CB * vector CD = 0, we can find the circumcircle radius r of triangle ABC
- 18. In rectangle ABCD, ab = 2, ad = 1, then vector AC × vector CD
- 19. In the diamond ABCD with side length of 1, ∠ bad = 60 ° and E is the midpoint of BC, then the vector AE is multiplied by the vector AC
- 20. Given that the side length of square ABCD is 2 and E is the midpoint of CD, what is the product of vector BD? And vector AE?